6. A random sample of 16 statistics exams was taken. The average score in the sample was 76.2 with avariance of 144. The 99% confidence interval for the true average exam score isA. 49.677 to 102.723C. 67.359 to 85.041B. 73.99 to 78.41D. 67.437 to 84.963

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Answered: 6. A random sample of 16 statistics…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Insurance Company A claims that its customers pay less for car insurance, on average, than customers of its competitor, Company B. You wonder if this is true, so you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 13 people who buy insurance from Company A, the mean cost is $⁢150 per month with a standard deviation of $⁢19. For 9 randomly selected customers of Company B, you find that they pay a mean of $⁢157 per month with a standard deviation of $16. Assume that both populations are approximately normal and that the population variances are equal to test Company A’s claim at the 0.05 level of significance. Let customers of Company A be Population 1 and let customers of Company B be Population 2. Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places. Step 3 of 3: Draw a conclusion and interpret the decision.
Insurance Company A claims that its customers pay less for car insurance, on average than customers of its competitor, Company B. You wonder if this is true, so you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 11 people who buy insurance from Company A, the mean cost is $⁢150 per month with a standard deviation of $⁢14. For 14 randomly selected customers of Company B, you find that they pay a mean of $⁢158 per month with a standard deviation of $⁢12. Assume that both populations are approximately normal and that the population variances are equal to test Company A’s claim at the 0.10 level of significance. Let customers of Company A be Population 1 and let customers of Company B be Population 2. Step 2 of 3 : Compute the value of the test statistic. Round your answer to three decimal places.
For a study conducted by the research department of a pharmaceutical company, 235 randomly selected individuals were asked to report the amount of money they spend annually on prescription allergy relief medication. The sample mean was found to be $17.80 with a standard deviation of $4.30 . A random sample of 250 individuals was selected independently of the first sample. These individuals reported their annual spending on non-prescription allergy relief medication. The mean of the second sample was found to be $18.20 with a standard deviation of $4.00 . As the sample sizes were quite large, it was assumed that the respective population standard deviations of the spending for prescription and non-prescription allergy relief medication could be estimated as the respective sample standard deviation values given above. Construct a 90% confidence interval for the difference −μ1μ2 between the mean spending on prescription allergy relief medication ( μ1 ) and the mean…
Insurance Company A claims that its customers pay less for car insurance, on average, than customers of its competitor, Company B. You wonder if this is true, so you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 13 people who buy insurance from Company A, the mean cost is $150 per month with a standard deviation of $19. For 9 randomly selected customers of Company B, you find that they pay a mean of $157 per month with a standard deviation of $16. Assume that both populations are approximately normal and that the population variances are equal to test Company A's claim at the 0.05 level of significance. Let customers of Company A be Population 1 and let customers of Company B be Population 2. Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places.
In a simple random sample of 144 households in a county in Virginia, the average number of children in these households was 3.62 children. The standard deviation from this sample was 2.40 children. A 90% confidence interval for the mean number of children in all households in this county is?
a. 14.0, 16.3, 18.7, 19.9, 21.3 b. 16.1, 13.4, 13.6, 14.9, 14.1, 10.5 Construct the 95% confidence interval for the difference in u of the two given samples.
According to the College Board, scores on the math section of the SAT Reasoning college entrance test for the class of 2010 had a mean of 516 and a standard deviation of 116. Assume that they are roughly normal.One of the quartiles of the scores from the math section of the SAT Reasoning test is 438. The other quartile is _______.
A simple random sample of 5 observations from a normal distribution was taken, and the following values were obtained: {15, 19, 19, 20, 22}. A 95% confidence interval for the mean is O A. (16.569, 21.431) O B. (15.835, 22.165) O C.(16.069, 21.931) O D. (17.124, 20.876) O E. (16.765, 21.235)
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Suppose that Randy is an analyst for the bicyling industry and wants to estimate the asking price of used entry-level road bikes advertised online in the southeastern part of the United States. He obtains a random sample of ?=11 online advertisements of entry-level road bikes. He determines that the mean price for these 11 bikes is ?⎯⎯⎯=$703.75 and that the sample standard deviation is ?=$189.56. He uses this information to construct a 95% confidence interval for ?, the mean price of a used road bike. a) What is the lower limit of this confidence interval? Please give your answer to the nearest cent. b) What is the upper limit of this confidence interval? Please give your answer to the nearest cent.

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6. A random sample of 16 statistics exams was taken. The average score in the sample was 76.2 with a variance of 144. The 99% confidence interval for the true average exam score is A. 49.677 to 102.723 C. 67.359 to 85.041 B. 73.99 to 78.41 D. 67.437 to 84.963